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PRACTICE FINAL SOLUTIONS 1. Tangent line x = 6 + 2t, y = −2 − t, z = 9 + 6t, perpendicular plane 2(x − 6) − (y + 2) + 6(z − 9) = 0 2. See Graphs at end. The surface is a hyperboloid of one sheet. The graph of the surface is pretty much exactly Figure 8 on pg. 605 of your book. 3. Approach along y-axis = −1, Approach along x-axis = 1 √ 29 < −4, 8 >, rate of increase in other direction = 4. Fastest increase direction = 1/ √ 32/ 29 5. −1(x + 2) + 2(y − 1) − 2/3(z + 3) = 0 6. Local max achieved along the line y = x, global min of −1 achieved √ (and global) √ √ of 1 √ at ( 2/2, − 2/2), (− 2/2, 2/2) 7. 44 8. 3π/2 9. 1/3(e3 − 1) 10. 16/5π 11. Graph included at end. Note that the vectors corresponding to points x = 0 do not exist because of division by 0. 12. div = 1, curl = 0 13. 320 14. −59/105 15. Potential function f (x, y) = x2 cos y − y sin x + C, integral = π 2 16. Potential function f (x, y) = 3x2 + 5xy + 2y 2 + C, integral = 0 17. e − 1 18. The first calculation represents flux of the vector field through the unit circle, and is 2π. The second calculation represents circulation (or rotation) of the vector field around the unit circle, and is 0. The fact that this is 0 is a result of the fact that the vector field is orthogonal to the unit circle at each point, so there is no tendency to rotate. √ √ 19. Surface area = π/6(17 17 − 1), Mass = π/60(391 17 + 1) 1 20. −4/3π 21. π/2 22. 0 2